Does anyone know the math for finding out your flywheel HP & TQ versus your rear wheel power?

 

Its about a 16% drivetrain loss thru a 6 speed but then I see you have a 12 bolt so maybe somewhere between 18-20%.

 

I know about the % loss and all but isn't there a specific formula I can use? I saw a chart at a dyno shop one time but don't remember it.

 

I am not familiar with any formula. When I dynoed my car the owner just multiplied my RWHP by 1.16 to get the power at the FW.

 

Haha thanks TurboZ, thats the math I was looking for. I put down 414 at the wheels so according to that math its 480.24 at the flywheel.

 

You divide it by the number, not multiply. Say its 16% loss, you divide your RWHP by .84 to get the FW #'s.

 

Thanks guys. With both formulas I'm between 480-487FWHP. Nows heres a question, whats up with my TQ? I dynoed 414 RWHP but only 350 RWTQ?? That is a N/A pull, shouldn't it be a little higher or is that common?

 

What CI motor? The TQ is just about right for a 350. I made the same HP as you with my 383 but made 412 Ft Pnd of TQ. The whole reason that I went with bigger cubes was to pick up TQ across the board. I make 350 ftpnds at 1800 RPM and it peaked at 4500 with 412 but held above 300 till 6500 where I quit making HP as well. I realy think there is more in my combo considering I had a miss the day I dynoed and still have stock lifters (New but stock)

 

Its still a 350ci Ej, so that sounds right?

 

I was able to measure my engine's performance on an engine dyno with full intake and exhaust, and then in the car on a chassis dyno. At the 500flywheelHP level, the drivetrain losses were only 12.6% for:

-steel flywheel Street Twin
-stock T56
-3" chrome moly DS (19#)
-Strange 12-bolt with 3.73's
-275/40-17's on OZ Monte Carlos.

I have never seen any evidence that the T56 loses 16% or 18-20% with a 12-bolt.

If you believe my numbers, you can either divide by 0.874

414 / 0.874 = 474 flywheel

or you can multiply by 1.1442

414 x 1.1442 = 474 flywheel

 

Superflow says there is no formula but I have always figured "about" 10% for stick and 20% for auto.That will get you in the ballpark.

 

Multiplication and division are opposites, so you can divide by .84 (1.00 - 0.16) or multiply by 1.16 (1.00 + 0.16). Everyone has their own opinions on what percentage you lose through the drivetrain. For a T-56, I'd lean closer to Injuneer's 12.6 % than 16-18%.

 

No they aren't, and no you can't. If you accept that the losses are 16%, you can either divide rwHP by 0.84 [1.00 - 0.16 = 0.84], or you can multiply rwHP by 1.1905 [ = 1 / (1.00 - 0.16) = 1 / 0.84 = 1.1905].

414 / 0.84 = 493 flywheel

414 X 1.1905 = 493 flywheel

 

Man--- that sounds like a big gain(or loss) for a stick--figuring it at 16%.Thats more like an auto with a converter and a 9".16% must be way high. The 10-12% sounds more reasonable to me but I have not dynoed every car on the planet,just a few.

 

Wow, people are throwing out some crazy numbers here, and how to calculate fwhp!

Injuneer I believe since he has the data.

PANZ28,

It's pretty simple to figure out your drivetrain loss. You dynoed 253 rwhp. The LT1 is rated at 285 fwhp. (If GM rated them correctly).

253/285 = 0.89
So
1.0-0.89 = 11% drivetrain loss.

Take any rwhp numbers that you have and divide by 0.89 to get the flywheel numbers. So for you, 414/0.89 = 465 fwhp

I've always used 11% drivetrain loss for the M6 to calculate fwhp.

Dan